Cauchy Distribution does not have a mean, but
\(\int^\infty_0{x.f(x)}dx=\int^\infty_0{\cfrac{2x^2}{\pi(1+x^4)}}dx=\cfrac{1}{\sqrt{2}}\)
when we consider both \(-\theta\) and \(\theta\)
\(E\left\{x\right\}=2\cfrac{1}{\sqrt{2}}=\sqrt{2}\)
The emitted photons, assuming that \(\theta\) is uniformly distributed, \(-\infty\lt\theta\lt\infty\), has an expected value of \(\sqrt{2}\).
But,
\(\int{x^2.f(x)}dx=\int{\cfrac{2x^3}{\pi(1+x^4)}}dx=\cfrac{1}{2\pi}ln(x^4+1)\)
does not converge for all values of \(\theta\), however for the range \(0\le x\le \cfrac{\pi}{2}\),
\(E\left\{x^2\right\}=0.31169\)
in which case, over the same range,
\(E\left\{x\right\}=0.314044\)
and the variance is,
\(Var\left\{x\right\}=2*0.31169-(2*0.314044)^2\)
\(Var\left\{x\right\}=0.22889\), \(\sigma=\sqrt{Var\left\{x\right\}}=0.4784\)
Polarization mean and variance has little meaning, compared to mode at \(\cfrac{1}{\sqrt[4]{3}}\).
